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Grade: 12

                        

A cup of tea cool from 80 degree Celsius to 60 degree celsius in one minute tab indent temperature is is 60 degree in cooling from 60 degree Celsius 50 degree it will take what is the answer options 1) 30 sec 2) 60sec 3) 90sec 4) 50sec

one year ago

Answers : (2)

Vikas TU
13786 Points
							
Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] 
( Tf - Ti)/t = K[ ( 80 + 60)/2 - 60]
( 80-60)/1 = K[ 70 - 60]
20 = K× 10 
K = 20/10 = 2
in second condition, 
initial temperature ( Ti) = 60°C 
Final temperature ( Tf) = 50°C 
Time taken for cooling is t 
A/C Newton's law of cooling 
( 60 - 50)/t = 2 [ (60+50)/2 -60]
10/t = 2* 5 
10/t = 10  
t = 1 min.
one year ago
Khimraj
3007 Points
							
According to Rate of cooling
dT/dt  = K((T_initial+T_final)/2 - T_surr) 
( 80-60)/1 = K(70-60)
K = 2
Time taken for cooling
(60 - 50)/t = 2 ((60+50)/2 -60)
t = 60 sec
one year ago
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